SAT Question 4c774b00: Answer, Explanation & Desmos Solution

Ages of 20 Students Enrolled in a College Class
Age	Frequency
$18$	$6$
$19$	$5$
$20$	$4$
$21$	$2$
$22$	$1$
$23$	$1$
$30$	$1$

The table above shows the distribution of ages of the $20$ students enrolled in a college class. Which of the following gives the correct order of the mean, median, and mode of the ages?

Difficulty:Medium

$mode < median < mean$

$mode < mean < median$

$median < mode < mean$

$mean < mode < median$

Explanation

Correct answer: A.

Choice A is correct. The mode is the data value with the highest frequency. So for the data shown, the mode is $18$ . The median is the middle data value when the data values are sorted from least to greatest. Since there are $20$ ages ordered, the median is the average of the two middle values, the $10$ th and $11$ th, which for these data are both $19$ . Therefore, the median is $19$ . The mean is the sum of the data values divided by the number of the data values. So for these data, the mean is $\frac{( 18 \times 6 ) + ( 19 \times 5 ) + ( 20 \times 4 ) + ( 21 \times 2 ) + ( 22 \times 1 ) + ( 23 \times 1 ) + ( 30 \times 1 )}{20} = 20$ . Since the mode is $18$ , the median is $19$ , and the mean is $20$ , mode $<$ median $<$ mean.

Desmos solution:

Note: The table's most frequent age is 18, and the calculator shows the median 19 and mean 20.

Enter in Desmos:

$\frac{19 + 19}{2}$
$\frac{18 ( 6 ) + 19 ( 5 ) + 20 ( 4 ) + 21 ( 2 ) + 22 ( 1 ) + 23 ( 1 ) + 30 ( 1 )}{20}$

Reviewed and rewritten by CookSAT

Why the other choices are wrong

Choice B

Choice B is incorrect because the mean is greater than the median. Alternate approach: After determining the mode, $18$ , and the median, $19$ , it remains to determine whether the mean is less than $19$ or more than $19$ . Because the mean is a balancing point, there is as much deviation below the mean as above the mean. It is possible to compare the data to $19$ to determine the balance of deviation above and below the mean. There is a total deviation of only $6$ below $19$ (the $6$ values of $18$ ); however, the data value $30$ alone deviates by $11$ above $19$ . Thus the mean must be greater than $19$ .

Choice C

Choice C is incorrect because the median is greater than the mode. Alternate approach: After determining the mode, $18$ , and the median, $19$ , it remains to determine whether the mean is less than $19$ or more than $19$ . Because the mean is a balancing point, there is as much deviation below the mean as above the mean. It is possible to compare the data to $19$ to determine the balance of deviation above and below the mean. There is a total deviation of only $6$ below $19$ (the $6$ values of $18$ ); however, the data value $30$ alone deviates by $11$ above $19$ . Thus the mean must be greater than $19$ .

Choice D

Choice D is incorrect because the mean is greater than the median. Alternate approach: After determining the mode, $18$ , and the median, $19$ , it remains to determine whether the mean is less than $19$ or more than $19$ . Because the mean is a balancing point, there is as much deviation below the mean as above the mean. It is possible to compare the data to $19$ to determine the balance of deviation above and below the mean. There is a total deviation of only $6$ below $19$ (the $6$ values of $18$ ); however, the data value $30$ alone deviates by $11$ above $19$ . Thus the mean must be greater than $19$ .

SAT Question 4c774b00: Distributions, Center & Spread answer and explanation with Desmos solution

Explanation

Enter in Desmos:

Why the other choices are wrong